Advertisements
Advertisements
प्रश्न
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
विकल्प
zero
`-(0.45 hati + 0.6 hatj)`
`-(0.9 hati + 1.2 hatj)`
`-5 (hati + hatj)`
Advertisements
उत्तर
`-(0.9 hati + 1.2 hatj)`
Explanation:
Given, `u = (3hati + 4hatj)` m/s
And `v = - (3hati + 4hatj)` m/s
Mass of the ball = 150 g = 0.15 kg
Δp = Change in momentum
= Final momentum – Initial momentum
= `mv - mu`
= `m(v - u) = (0.15) [- (3hati + 4hatj) - (3hati + 4hatj)]`
= `(0.15) xx [ - 6hati - 8hatj]`
= `- [0.15 xx 6hati + 0.15 xx 8hatj]`
= `- [0.9 hati + 1.20 hatj]`
Hence, Δp = `-[0.9 hati + 1.2 hatj]`
APPEARS IN
संबंधित प्रश्न
Explain why a cricketer moves his hands backwards while holding a catch.
A car accelerates on a horizontal road due to the force exerted by.
A person says that he measured the acceleration of a particle to be non-zero even though no force was acting on the particle.
Two blocks A and B of mass mA and mB , respectively, are kept in contact on a frictionless table. The experimenter pushes block A from behind, so that the blocks accelerate. If block A exerts force F on block B, what is the force exerted by the experimenter on block A?
A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg, respectively. Assuming that the magnitudes of acceleration and deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2.
Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s2. Find the elongation.
Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in the following figure. Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?
Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.
In the following figure shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling by a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight recorded on the machine? What force should he exert on the rope to record his correct weight on the machine?
A body of mass m moving with a velocity v is acted upon by a force. Write an expression for change in momentum in each of the following cases: (i) When v << c, (ii) When v → c and (iii) When v << c but m does not remain constant. Here, c is the speed of light.
State the Newton's second law of motion. What information do you get from it?
Use Newton's second law of motion to explain the following instance :
An athlete prefers to land on sand instead of hard floor while taking a high jump .
The linear momentum of a ball of mass 50 g is 0.5 kg m s-1. Find its velocity.
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : The magnitude of the force
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
State two factors which determine the momentum of a body.
State Newton's second law of motion. Is Newton's first law of motion contained in Newton's second law of motion?
A stone is dropped from a cliff 98 m high.
What will be its speed when it strikes the ground?
A metre scale is moving with uniform velocity. This implies ______.
The position time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.
